Hausdorff question

[Nicoco]

Does anyone have an idea how to prove the following:

 

[/url]

[math] h(A\cup B, C\cup D) \leq \max\{h(A,C), h(B,D)\} [/math]

 

with the Hausdorff distance. I can see it on a drawing, but I'm not able to prove it correctly.

 

Thanks in advance

[Dapthar]

I'm working on it, and I'll post a reply in a few minutes.

[Dapthar]

Does anyone have an idea how to prove the following:

 

[math] h(A\cup B' date=' C\cup D) \leq \max\{h(A,C), h(B,D)\} [/math']

(I think that there might be a small problem with your notation, since if the sets in question are open, there is, in general, no maximum or minimum distance between them. You would have to use [math]\sup[/math] and [math]\inf[/math] instead of [math]\min[/math] and [math]\max[/math] respectively.)

 

As with most things in Topology, it really helps to have an intuitive notion of what a definition 'actually means'. In this case, here's what I have as my 'internal definition' of [math]h(A,B) = r[/math]

 

"Every point in [math]A[/math] is at most [math]r[/math] units away from a point in [math]B[/math]."

 

Thus, with this in mind, we can formulate an idea for the proof. (You should be able to handle the details on your own.)

 

So, what does [math]h(A\cup B, C\cup D) = r_1[/math] mean? It means that:

 

Every point in [math]A[/math] and [math]B[/math] is at most [math]r_1[/math] units away from some point in [math]C[/math] or [math]D[/math].

 

With this in mind, we can easily see that (i.e. you have to manipulate the definitions to prove this):

 

[math]h(A\cup B, C\cup D) \leq \max\{h(A, C\cup D),h(B,C\cup D)\}[/math].

 

However, it is also clear that (again, fill in the details):

 

[math]\max\{h(A, C\cup D),h(B,C\cup D)\} \leq[/math][math] \max\{h(A, C), h(B,D)\} [/math] (This part is simply a basic application of the definition of Hausdorff distance.)

 

Technically, [math] h(A\cup B, C\cup D) \leq \max\{h(A,D), h(B,C)\} [/math] as well, since the problem is 'symmetric with respect to the sets'.

[matt grime]

He, the OP is probably talking of the Hausdorff distance between a finite set of points, at which point it makes sense to talk of mins and maxs

[Nicoco]

sorry, double posted this...

[Nicoco]

Yes, it has to be with and instead of the and . I mixed up two definitions.

 

By the way, in the meantime I was able to prove it in a different (but somewhat more tricky) way, you might be interested.

 

First, I give the definition of the -thickening of a set :

is the collection of all points within a distance of a point of A.

It's similar to a Ball in a metric space.

 

Next, let and . Then it is easy to see that

 

Take , then is easely follows that

 

Hence,

It is easy to show that

 

Combining these last two inclusions we see

You can prove, using the same arguments, that

 

And so it follows that,

 

But your approach is a tad more easy I think...